I'm trying to solve the following differential equation:
$y''(t)+3y'(t)+2y(t) = f(t)$ , where $f(t) =(t, 0 <= t <= 1), (0, 1< t) $ with initial condition being $y(0)=0=y'(0)$.
Applying the Laplace Transform to both sides I obtain
$F(p) = (1-e^(-p))/(p^2(p+1)(p+2))$ with $Re(p)>0$.
How should I continue?
Take the partial fraction of $$\frac{1}{t^2\left(t+1\right)\left(t+2\right)}=\frac{1}{2t^2}+\frac{1}{t+1}-\frac{1}{4\left(t+2\right)}-\frac{3}{4t}.$$ \begin{align*} \mathscr{L}(y)= & \frac{1-e^{-t}}{2t^2}+\frac{1-e^{-t}}{t+1}-\frac{1-e^{-t}}{4\left(t+2\right)}-\frac{3(1-e^{-t})}{4t}\\ \implies y= & \mathscr{L}^{-1}\left( \frac{1-e^{-t}}{2t^2}+\frac{1-e^{-t}}{t+1}-\frac{1-e^{-t}}{4\left(t+2\right)}-\frac{3(1-e^{-t})}{4t} \right) \end{align*} Now use the Heaviside function for get the inverse. $$ \mathscr{L}^{-1}\left( \frac{e^{-t}}{t^2} \right)=\text{H}\left(t-1\right)\left(t-1\right) $$ So, $$ \mathscr{L}^{-1}\left( \frac{1-e^{-t}}{2t^2} \right)=\frac{t}{2}-\frac{1}{2}\text{H}\left(t-1\right)\left(t-1\right) $$ Similarly, the remaining can be computed.